Some Examples of Static Equilibrium — Question 1

Section 12-4 · Physics pages 1832–1833

Question 1 A ladder load is limited by friction before reaching the top A uniform 6 m ladder of mass 20 kg leans at 60◦above horizontal against a smooth wall. The floor has µs = .40. A 60 kg person stands distance s along the ladder from its foot. Use g = 10. person Ladder side view; the wall is smooth and only the floor supplies friction. (a) Find the largest admissible s before slipping. (b) Find floor and wall forces at that limit and check that the person is still on the ladder.
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Question 1 - Solution A ladder load is limited by friction before reaching the top person (a) Derivation and calculation Let H be the wall reaction and f the floor friction. Force balance gives N = (20 + 60)10 = 800 N, f = H. At impending slip, H = µsN = .40(800) = 320 N. Moments about the foot give HL sin 60◦= ( 200L 2 + 600s ) cos 60◦. (b) Result and interpretation Solving with L = 6.0 m, smax = 320(6) tan 60◦−200(3) 600 = 4.54 m . The limiting reactions are N = 800 N , H = f = 320 N . Since smax < L, slip limits the climb before the top is reached. Physical check The floor friction opposes outward foot motion and the smooth wall supplies only a horizontal reaction.

Question 1

A ladder load is limited by friction before reaching the top

A uniform 6m6\,\mathrm m ladder of mass 20kg20\,\mathrm{kg} leans at 60∘60^\circ above horizontal against a smooth wall. The floor has μs=.40\mu_s=.40. A 60kg60\,\mathrm{kg} person stands distance ss along the ladder from its foot. Use g=10g=10.

Diagram shown in the original SVG worksheet.

  1. Find the largest admissible ss before slipping.

  2. Find floor and wall forces at that limit and check that the person is still on the ladder.

Question 1 - Solution

A ladder load is limited by friction before reaching the top

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Let HH be the wall reaction and ff the floor friction. Force balance gives N=(20+60)10=800N,f=H.N=(20+60)10=800\ \mathrm N,\qquad f=H. At impending slip, H=μsN=.40(800)=320H=\mu_sN=.40(800)=320 N. Moments about the foot give HLsin⁡60∘=(200L2+600s)cos⁡60∘.HL\sin60^\circ=\left(200\frac L2+600s\right)\cos60^\circ.

(b) Result and interpretation

Solving with L=6.0L=6.0 m, smax=320(6)tan⁡60∘−200(3)600=4.54m.s_{\max}=\frac{320(6)\tan60^\circ-200(3)}{600} =\boxed{4.54\ \mathrm m}. The limiting reactions are N=800N,H=f=320N.\boxed{N=800\ \mathrm N},\qquad\boxed{H=f=320\ \mathrm N}. Since smax<Ls_{\max}<L, slip limits the climb before the top is reached.

Physical check

The floor friction opposes outward foot motion and the smooth wall supplies only a horizontal reaction.